Dear Friends
Recently I have published an article related the imaginary parts of the zeros of the Riemann zeta function . http://ijmcs.future-in-tech.net/13.1/R-Gunes.pdf
Here , we have found a new formula which is very simple formula ( equation 3.3 in page 3).
Then, we have decided to find the imaginary parts of the zeros of the Riemann zeta function by this equation 6.1 in page 7.
Now, can you arrange a computer calculation program which will allow us will see the calculations with 500 decimal digits accuracy.
Thus if we use the calculation results of Prof Odlyzko about the zeros of the Riemann zeta function in this calculation program, can we see that the Riemann zeta function will be equal zero with 500 or 1000 decimal digits accuracy or not ?.
Shortly we can say here, without using Riemann-Siegel Function Z(t) , we can find the zeros of the Riemann zeta function.
Maybe, we can make just 6 calculations. These 6 calculations will be enough to see the sensitivity of this new formula .
I’m almost certain that Mathematica can do real calculations with that sort of precision but unfortunately I don’t yet know how to set the precision for such calculations.
I am more familiar with Number Theory Library (NTL) which does arbitrary precision arithmetic with both integers and reals. NTL does not do symbolic math though, so for instance you.d have to do the integrals on the RHSs by hand (or a symbolic math program) and then code it in C++, calling NTL where needed. FWIW, NTL is GPL3.
There’s also Axiom, for which there is a TeXmacs plugin. Axiom documentation is Huge. Finding how to set the precision may be difficult.
Best of luck,and contact me here if you want me to elaborate on any of this. Regards, Graham.
I have not read your paper but I assume that you are asking how to get 500 digits of accuracy in your calculation. If this is correct, you should use WorkingPrecision and decimal numbers with 500 digit accuracy. For example, 0.1 has only 16 digits of accuracy but
Dear Nail , Dear Graham
Dear Friend
Could you help me ? I want to see the following calculation results with the ´500´ decimal digits ( I have no any idea about computer programming)
Thank you very much for your this help.
s_1=0.5+it_1 and s_2=0.5+it_2
See t_1 and t_2 on below as with the ´500´ decimal digits .
Sorry to have taken so long to get back to you. I think I will have a bit more time for this from here on. I’ve noticed a few oddities and I’m afraid that I have some bad news.
The bad news first: the sum, counting up to N, for the large values of N that you want, is going to take too long for a single machine, especially since it involves a reciprocal and a power. Consider this: if incrementing n takes 1 clock; testing it against N another clock, raising n to the s’th another two clocks, taking the reciprocal another clock and adding the result to an accumulator yet another clock - that’s six clocks per iteration, assuming we’re using ordinary ints and doubles. A fast machine these days doesn’t run much faster than 4GHz, so our loop takes 1.5e-9 seconds per iteration. For your lowest N, 1.0e14, it would take 1.5e5 seconds, or about 2500 minutes, or 41 hours 40 minutes. You can expect the times to be hundreds of times worse for 500-digit precision. It will also be hundreds of times worse for 1.0e16, ditto for N==1.0e18:-(
Cloud computing might just save you here, and I understand that Mathematica can do some cloudy stuff.
I noticed that you originally had a graphic containing two equations (real & imaginary parts?) in your first post here; now it’s changed to rendered postscript or something. Anyhow, as I recall it there was an integral expression on the right. Now there’s a dx on the right which doesn’t seem to refer to anything.
